<span>a = (v2 - v1)/t= acceleration formula
a = (70 - 0)/7
a = 10 km/hr/sec
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It's better to use as few units as possible.
10 km/hr = 10 km*1000 m/km/(1 hr*3600 sec/hr) = 25/9 m/sec
a= 25/9 m/sec/se</span>
Complete Question
A proton is located at <3 x 10^{-10}, -5*10^{-10} , -5*10^{-10}> m. What is r, the vector from the origin to the location of the proton
Answer:
The vector position is 
Explanation:
From the question we are told that
The position of the proton is
Generally the vector location of the proton is mathematically represented as

So substituting values

<u>Answer:</u>
The amount of the lighted side of the moon you can see is the same during "how much of the sunlit side of the moon faces Earth".
<u>Explanation:</u>
The Moon is in sequential rotation with Earth, and thus displays the Sun, the close side, always on the same side. Thanks to libration, Earth can display slightly greater than half (nearly 59 per cent) of the entire lunar surface.
The side of the Moon facing Earth is considered the near side, and the far side is called the reverse. The far side is often referred to as the "dark side" inaccurately but it is actually highlighted as often as the near side: once every 29.5 Earth days. During the New Moon the near side becomes blurred.
To solve this problem we will apply the concepts related to Ohm's law and Electric Power. By Ohm's law we know that resistance is equivalent to,

Here,
V = Voltage
I = Current
While the power is equivalent to the product between the current and the voltage, thus solving for the current we have,


Applying Ohm's law


Therefore the equivalent resistance of the light string is 
Answer:
7.72 Liters
Explanation:
normal body temperature = T_body =37° C
temperature of ice water = T_ice =0°c
specfic heat of water = c_{water} =4186J/kg.°C
if the person drink 1 liter of cold water mass of water is = m = 1.0kg
heat lost by body is Qwater =mc_{water} ΔT
= mc{water} ( T_ice - T_body)
= 1.0×4186× (0 -37)
= -154.882 ×10^3 J
here negative sign indicates the energy lost by body in metabolic process energy expended due to brisk - hour long walk is Q_{walk} = 286 kilocalories
= 286×4186J
so number of liters of ice water have to drink is
n×Q_{water} =Q_{walk} n= Q_{walk}/ Q_{water}
= 286×4186J/154.882×10^3 J
= 7.72 Liters